NAME:
Midterm Exam 2
Chemistry 211
Oct. 14,1997
PAGE 1 0f 5
- Show all work or NO CREDIT will be awarded.
- Work problems to proper number of significant figures.
- Be sure to include units in your answers.
- Circle your final answers.
- Put your name on ALL pages.
(4 pts.) 1. Plotted below are graphs of heat absorbed versus temperature for two systems
(A & B). Which system has the larger heat capacity?
a) System A
b) System B
c) Both the same.
d) Not enough information to tell.
(4 pts.) 2. To a solution of 10 mL of 6 M NaOH is added enough water to make 100 mL
of solution. The number of moles of NaOH present is:
a) increased by a factor of 10
b) decreased by a factor of 10
c) unchanged
The concentration of NaOH is:
10
a) increased by a factor of 10
b) decreased by a factor of
c) unchanged
(4 pts.) 3. Solutions of NaNO3 and K2 SO4 are dissolved in water and the water then
allowed to evaporate. How many possible salts could be recovered?
a) 1
b) 2
c) 3
d) 4
e) Depends on the electrolytic potential.
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(4 pts.) 4. Given the thermochemical equation:
CH4(g) + 2 O2(g) -----> CO2(g) + H2 O(g) ……….Δ H˚rxn = -802 kJ
What amount of energy is released when 32.0 g of methane is burned?
a) 401 kJ
b) 802 kJ
c) 1604 kJ
d) cannot tell
(4 pts.) 5. When C(s,graphite) is burned to yield CO2 , 394 kJ of energy are released per mole
of C atoms burned. When C6 0 (Buckyball) is burned to yield CO2 approximately
435 kJ of energy is released per mole of carbon atoms burned. Would the
buckyball-to-graphite conversion be
a) exothermic
b) endothermic
c) thermoneutral
d) an extensive property
(4 pts.) 6. 100 mL of a 0.20M solution of HNO3
a) will conduct electricity better than 200 mL of a 0.15M solution of NH3 .
b) contains 0.060 moles of NO3 c) can be neutalized by the addition of 100 mL of 0.20M KMnO4 .
d) is basic.
(4 pts.) 7. A standard is prepared for the analysis of 2',3'-didehydro-3' deoxythymidine
(d4T) (C1 0H1 2N2 O4 , FW 224.2), an inhibitor of HIV. A stock solution is prepared
by dissolving 22.4 mg of d4T in 50.0 mL of solution. A 10.0 mL aliquot (portion)
of this solution is diluted to a final volume of 100.0 mL. The concentration of the
final solution is
a) 2.00 X 10-3 M
b) 2.00 X 10-4 M
c) 2.00 X 10-5 M
d) 5.00 X 10-4 M
e) 5.00 X 10-5 M
(4 pts.) 8. Which of the following is (are) weak bases?
a) KOH
b) NH3
c) NaCl
d) Ca(OH)2
2
(4 pts.) 9. Which of the following is not a unit of energy?
a) electron volt
b) calorie
c) Joule
d) Pascal
(4 pts.) 10. Based on the equations below, which metal is the most active?
Pb(NO3 )2 (aq) + Ni (s) ---------> Ni(NO3 )2 (aq) + Pb(s)
Pb(NO3 )2 (aq) + Ag (s) ---------> No reaction
Cu(NO3 )2 (aq) + Ag (s) --------> No reaction
a) Ni
b) Ag
c) Cu
d) Pb
11. Briefly (only the first 15 words will be graded) define the following terms:
(5 pts.) a. State Function – Depends only on initial and final states, not on path.
(5 pts.) b. Standard State - The physical state of an element or compound at
standard temperature and pressure.
(5 pts.) c. The First Law of Thermodynamics - The total energy of the universe is
constant.
(5 pts.) d. Oxidizing Agent - The substance that accepts electrons and is reduced
in a re-dox recation.
(5 pts.) 12. Under what condition(s) is the enthalpy change(ΔH) of a chemical reaction
equal to the heat of the reaction(q). Constant pressure.
(5 pts.) 13. How many mL of 0.35M HBr are needed to neutralize 52 mL of 0.035M
NaOH?
HBr + NaOH -------> NaBr + H2 O
M AVA = M BVB
M V
(0.035 M NaOH )(52mL)
VA = B B =
= 5.2mL
MA
0.35 M HBr
€
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14. Galactose is a common sugar very similar to glucose and is found in milk. If you do not
have the correct enzyme ( a genetic disorder easily tested for in infants) to convert galactose to
glucose you have to have restricted milk intake as a baby, or serious brain damage may result.
Suppose you have 12.5 grams of galactose (C6 H1 2O6 ) in 5.0 liters of solution.
(5 pts.) a. What is the molarity of your sugar solution?
 1mol C 6 H 12 O6 
12.5g x 
 = 0.0694 mol
180g


0.0694 mol
Molarity( M ) =
= 0.014 M
5.0L
€ pts.) b. If you drink 250. mL of the solution how many grams of galactose will you have
(5
ingested?



mols 
180g
0.014
 x (0.025L) x
 = 0.63g

L 
1mol C 6 H 12 O6 
(12 pts.) 15. Determine ΔHo rxn for the following reaction:
€
5 PbO2 (s) + 4 P (s,w) ⇒ P4 O1 0 (s) + 5 Pb (s)
Substance ΔHo f (kJ/mol)
PbO2 (s)
P4 O1 0 (s)
-277
-2984
o
ΔHrxn
= ∑ ΔHf,o products − ∑ ΔHf,o reactants
kJ 
kJ 


 kJ   

 kJ  
o
(
)
(
)
(
)
ΔHrxn
= (1 mol )x -2884
+
5
mol
x
0
−
5
mol
x
-277
+
4
mol
x


 mol   

 0 mol  

mol 
mol 
o
ΔHrxn
= −1599 kJ
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(8 pts.) 16. When 182 grams of gold is added to 22.1 grams of water at a temperature of
25.0 o C the final temperature of the resulting mixture is 27.5 o C. If the heat
capacity of gold is 0.129 J/g•K, what was the initial temperature of the gold
sample? (heat capacity of water is 4.18 J/g•K)
Heat given off by metal => Heat absorbed by water
Heat absorbed by water = (mass of water) (specific heat of water) (change in
temperature of water)
Heat absorbed by water = (22.1 g) (4.18 J/g o C) (27.5 o C - 25.00 o C )
Heat absorbed by water = 231 J
How much heat did the metal sample lose?
The metal sample lost 231 J
Calculate the initial temperature of gold.
change in temperature = (heat loss of metal)/((mass of metal)(Specific heat))
Final Temp – Initial Temp = (heat loss of metal)/((mass of metal)(Specific heat))
Initial Temp = Final Temp - (heat loss of metal)/((mass of metal)(Specific heat))
Initial Temp = 27.5 o C – (-231 J)/(182 g * 0.129 J/g o C)= 37.3 o C
(20 pts.) EXTRA CREDIT) Suppose you have 0.90 g of clover leaves and you extract the
the oxalic acid (H2 C2 O4 ) from them into a small amount of water. (Remember, oxalic acid is
diprotic, i.e., it has two titratable protons.) This solution of oxalic acid is found to require
40.0 mL of 0.050 M NaOH for titration to the equivalence point.
a) Write the balanced equation for this titration.
H 2C 2 O4 + 2NaOH ⇒ 2H2 O + Na2 C 2 O4
b) Calculate the number of moles of base used.

mol 
-3
# mols = M BVB =  0.050
 x (0.0400 L ) = 2.0 x 10 mol NaOH

L 
c) Calculate the number of moles of acid in the solution.
€
 1 mol H 2C 2O 4 
-3
2.0 x 10 -3 mol NaOH x 
 = 1.0 x 10 mol H 2C 2O 4
 2 mol NaOH 
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d) Calculate the weight percentage of H2 C2 O4 in the leaves.

g 
-3
mass H 2C 2O 4 = MW x # mols = 90
 x (1.0 x 10 mol H 2C 2O 4 ) = 0.090 g H 2C 2O 4
 mol 
 0.090 g H 2C 2O 4 
weight % = 
 x 100 = 10%
 0.90 g Clover leaves 
€
€
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